Secondary 4 Pure Physics Preliminary Examination Paper 2

TuitionGoWhere Practice Paper - Pure Physics Secondary 4

Mark Scheme / Suggested Answers

Subject: Pure Physics
Level: Secondary 4
Paper: Preliminary Examination Practice Paper 2 (Version 2)
Total Marks: 80

Section A: Structured Questions [50 marks]

Question 1 [5 marks]

(a) To vary the current in the circuit / to vary the voltage across the filament lamp / to change the brightness of the lamp. [1]

(b) As current increases, the filament temperature increases. The increased lattice vibrations impede electron flow, increasing resistance. [2]
Accept: Higher current → higher temperature → higher resistance of metal filament.

(c) $R = \frac{V}{I} = \frac{6.0}{0.80} = = = 7.5 \Omega$ [1]

(d) Graph sketch:

• Axes labelled: Voltage / V (x-axis), Current / A (y-axis)
• Curve passes through origin (0,0) and point (6.0, 0.80)
• Curve has decreasing gradient (concave down) as voltage increases
• Shape shows non-ohmic behaviour [1]

Question 2 [6 marks]

(a) $\frac{V_s}{V_p} = \frac{N_s}{N_p} \Rightarrow \frac{12}{240} = \frac{N_s}{1200} \Rightarrow N_s = \frac{12 \times 1200}{240} = 60 \text{ turns}$ [2]

(b) For ideal transformer: $V_p I_p = V_s I_s$
With 90% efficiency: $0.90 \times V_p I_p = V_s I_s$
$0.90 \times 240 \times I_p = 12 \times 2.5$
$I_p = \frac{12 \times 2.5}{0.90 \times 240} = \frac{30}{216} = 0.139 \text{A}$ [2]

(c) To reduce eddy currents induced in the core, which would cause heating and energy loss. [1]

(d) Any one:

• Resistance of coils causes $I^2R$ heating losses
• Eddy currents in core cause heating
• Hysteresis loss in core during magnetisation cycles
• Flux leakage (not all flux links both coils) [1]

Question 3 [7 marks]

(a) $F = B I L \sin\theta = 0.50 \times 3.0 \times 0.080 \times \sin 90^\circ = 0.12 \text{N}$ [2]

(b) Direction: Into the page (or perpendicular to both current and field, determined by Fleming's Left Hand Rule: First finger = Field down, Second finger = Current left to right, Thumb = Force into page) [1]

(c) Maximum torque on coil: $\tau_{\text{max}} = N B I A$
Area $A = 0.080 \times 0.040 = 3.2 \times 10^{-3} \text{m}^2$
$\tau_{\text{max}} = 50 \times 0.50 \times 3.0 \times 3.2 \times 10^{-3} = 0.24 \text{Nm}$ [2]

(d) Torque $\tau = N B I A \sin\theta$, where $\theta$ is angle between coil normal and magnetic field. As coil rotates, $\theta$ changes, so $\sin\theta$ varies from 0 to 1 to 0, causing torque to vary. Maximum when coil plane parallel to field ($\theta = 90^\circ$), zero when coil plane perpendicular to field ($\theta = 0^\circ$). [2]

Question 4 [8 marks]

(a) $\mathcal{E}_0 = N B A \omega = N B A (2\pi f)$
$\mathcal{E}_0 = 200 \times 0.40 \times 1.5 \times 10^{-3} \times 2\pi \times 50 = 37.7 \text{V}$ [2]

(b) $V_{\text{rms}} = \frac{\mathcal{E}_0}{\sqrt{2}} = \frac{37.7}{\sqrt{2}} = 26.7 \text{V}$ [1]

(c) Graph sketch:

• Axes: Time / s (x-axis), Induced emf / V (y-axis)
• Sinusoidal wave starting at 0 V at t=0
• Period $T = \frac{1}{f} = \frac{1}{50} = 0.020 \text{s}$
• Two complete cycles shown (0 to 0.040 s)
• Peak amplitude = 37.7 V [2]

(d) Mean power $P_{\text{mean}} = \frac{V_{\text{rms}}^2}{R} = \frac{(26.7)^2}{10} = 71.3 \text{W}$ [2]

(e) Any one:

• Increase number of turns $N$
• Increase magnetic flux density $B$
• Increase coil area $A$ [1]

Question 5 [6 marks]

(a) Peak voltage $V_0 = 3.2 \text{ div} \times 5.0 \text{V/div} = 16.0 \text{V}$ [1]

(b) $V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{16.0}{\sqrt{2}} = 11.3 \text{V}$ [1]

(c) Period $T = 4.0 \text{ div} \times 2.0 \text{ms/div} = 8.0 \text{ms} = 8.0 \times 10^{-3} \text{s}$
Frequency $f = \frac{1}{T} = \frac{1}{8.0 \times 10^{-3}} = 125 \text{Hz}$ [2]

(d) Battery voltage $= 2.4 \text{ div} \times 5.0 \text{V/div} = 12.0 \text{V}$ [1]

(e) High input resistance draws negligible current from the circuit under test, so it does not alter the voltage being measured (minimises loading effect). [1]

Question 6 [6 marks]

(a) Initial flux linkage = 0 (magnet outside)
Final flux linkage $= N B A = 500 \times 0.30 \times 2.0 \times 10^{-3} = 0.30 \text{Wb turns}$
Change $= 0.30 \text{Wb turns}$ [2]

(b) Average emf $= \frac{\text{change in flux linkage}}{\text{time}} = \frac{0.30}{0.25} = 1.2 \text{V}$ [2]

(c) Direction: Anticlockwise (as viewed from magnet entry end).
Explanation: As N-pole approaches, magnetic flux through solenoid increases downwards. By Lenz's law, induced current opposes this increase by creating an upward magnetic field (N-pole at entry end). Using right-hand grip rule, this requires anticlockwise current as viewed from entry end. [2]

Question 7 [6 marks]

(a) $F = B I L \sin\theta = 0.60 \times 4.0 \times 0.12 \times \sin 30^\circ = 0.144 \text{N}$ [2]

(b) For a curved wire in uniform B-field, force equals force on straight wire connecting same endpoints.
Effective length = diameter $= 2 \times 0.060 = 0.12 \text{m}$ (perpendicular to field)
$F = B I L = 0.60 \times 4.0 \times 0.12 = 0.288 \text{N}$ [2]

(c) Magnetic force on a current element: $d\vec{F} = I d\vec{l} \times \vec{B}$. For uniform $\vec{B}$, total force $\vec{F} = I (\int d\vec{l}) \times \vec{B} = I \vec{L}_{\text{net}} \times \vec{B}$, where $\vec{L}_{\text{net}}$ is vector sum of all length elements = straight line from start to end. Since both wires have same endpoints and same current, net force is identical. [2]

Question 8 [6 marks]

(a) $I_{\text{kettle}} = \frac{P}{V} = \frac{2000}{240} = 8.33 \text{A}$
$I_{\text{iron}} = \frac{1500}{240} = 6.25 \text{A}$
Total current $= 8.33 + 6.25 = 14.58 \text{A}$ [2]

(b) Total current (14.58 A) > fuse rating (13 A). Fuse would blow under normal operation when both appliances are on. Not suitable – fuse rating should exceed maximum expected operating current. [2]

(c) Large current → strong magnetic field in electromagnet → attracts soft iron armature → releases latch/spring → opens contacts → breaks circuit. When current returns to normal, spring resets contacts. [2]

Section B: Longer Structured Questions [30 marks]

Question 9 [12 marks]

(a) $E = V + I r$ (or $V = E - I r$) [1]

(b) Graph:

• Points plotted accurately from table
• Best-fit straight line with negative gradient
• y-intercept at $V = 1.5 \text{V}$ (emf)
• x-intercept at $I = 1.5 \text{A}$ (short-circuit current)
• Axes labelled: Terminal p.d. V / V (vertical), Current I / A (horizontal) [3]

(c)(i) emf = y-intercept = 1.5 V [1]

(c)(ii) Gradient $= \frac{\Delta V}{\Delta I} = \frac{0.75 - 1.25}{0.75 - 0.25} = \frac{-0.50}{0.50} = -1.0$
Internal resistance $r = -\text{gradient} = 1.0 \Omega$ [2]

(d) Cell is old / depleted / has higher internal resistance due to chemical degradation / increased electrolyte resistance. [1]

(e) Maximum power delivered to load when $R_{\text{load}} = r = 1.0 \Omega$
$P_{\text{max}} = \frac{E^2}{4r} = \frac{(1.5)^2}{4 \times 1.0} = 0.5625 \text{W}$ [3]

(f) As current increases, voltage drop across internal resistance ($Ir$) increases. Since $V = E - Ir$, terminal p.d. decreases. Energy is dissipated as heat inside the cell. [2]

Question 10 [10 marks]

(a) Kinetic energy gained $= qV = \frac{1}{2}mv^2$
$v = \sqrt{\frac{2qV}{m}} = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 2000}{m}}$
But mass not given for this part – assume general derivation or use mass from (c).
Wait: Part (a) asks for speed of ion with charge +e accelerated through 2000 V. Mass not given here.
Correction: The question likely expects the expression or calculation using mass from (c). However, as written, mass is only given in (c).
Revised interpretation: Part (a) is general: $v = \sqrt{\frac{2eV}{m}}$. But without m, cannot compute numerical value.
Alternative: Perhaps the ion in (a) is the same as in (c). Then:
$v = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 2000}{3.3 \times 10^{-26}}} = \sqrt{1.939 \times 10^{10}} = 1.39 \times 10^5 \text{m/s}$ [2]

(b) Centripetal force = Magnetic force: $\frac{mv^2}{r} = Bqv \Rightarrow r = \frac{mv}{Bq}$ [1]

(c) Using $v$ from (a): $r = \frac{mv}{Bq} = \frac{3.3 \times 10^{-26} \times 1.39 \times 10^5}{0.25 \times 1.6 \times 10^{-19}} = 0.115 \text{m} = 11.5 \text{cm}$ [2]

(d) $r \propto m$ (since $v$ same for same charge and accelerating voltage). Ratio $= 2:1$ [1]

(e) In velocity selector: $qE = qvB \Rightarrow v = \frac{E}{B} = \frac{5000}{0.10} = 5.0 \times 10^4 \text{m/s}$ [2]

(f) Ions with $v > E/B$ experience magnetic force $>$ electric force → deflected one way. Ions with $v < E/B$ experience electric force $>$ magnetic force → deflected opposite way. Only $v = E/B$ gives zero net force. [2]

Question 11 [8 marks]

(a) Photon energy $E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{450 \times 10^{-9}} = 4.42 \times 10^{-19} \text{J}$
In eV: $\frac{4.42 \times 10^{-19}}{1.60 \times 10^{-19}} = 2.76 \text{eV}$ [2]

(b) $K_{\text{max}} = e V_s = 1.60 \times 10^{-19} \times 0.85 = 1.36 \times 10^{-19} \text{J}$ [1]

(c) Work function $\phi = E - K_{\text{max}} = 2.76 - 0.85 = 1.91 \text{eV}$ [2]

(d)(i) Maximum kinetic energy unchanged – depends only on frequency/wavelength, not intensity. [1]

(d)(ii) Photocurrent doubles – intensity doubled means twice as many photons per second, so twice as many photoelectrons emitted per second (assuming each photon ejects one electron). [1]

(e) As wavelength increases, photon energy decreases. When photon energy $<$ work function ($hc/\lambda < \phi$), no photoelectrons can be emitted regardless of intensity. Emission stops at threshold wavelength $\lambda_0 = hc/\phi$. [1]

End of Mark Scheme

Total: 80 marks